<h3>
Answer:</h3>
2.809 L of H₂SO₄
<h3>
Explanation:</h3>
Concept tested: Moles and Molarity
In this case we are give;
Mass of solid sodium hydroxide as 13.20 g
Molarity of H₂SO₄ as 0.235 M
We are required to determine the volume of H₂SO₄ required
<h3>First: We need to write the balanced equation for the reaction.</h3>
- The reaction between NaOH and H₂SO₄ is a neutralization reaction.
- The balanced equation for the reaction is;
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
<h3>Second: We calculate the umber of moles of NaOH used </h3>
- Number of moles = Mass ÷ Molar mass
- Molar mass of NaOH is 40.0 g/mol
Moles of NaOH = 13.20 g ÷ 40.0 g/mol
= 0.33 moles
<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
- From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
- Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
- Thus, Moles of H₂SO₄ = moles of NaOH × 2
= 0.33 moles × 2
= 0.66 moles of H₂SO₄
<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
- When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
- That is; Volume = Number of moles ÷ Molarity
In this case;
Volume of the acid = 0.66 moles ÷ 0.235 M
= 2.809 L
Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.
Evaporation, Condensation, precipitation and collection would be stages
Answer:
1000 µL; 10 µL
Explanation:
A p1000 micropipet is set to dispense 1000 µL.
A p10 micropipet set to dispense 10 µL.
Answer:
The thermodynamic parameter which is of significance in this case is the 'Reduction Potential' for molecular bromine which is ~ +1.1 v vs N.H.E. In other words, it is a strong oxidizing agent. The bromine will oxidize sulfur compounds in which the valence of sulfur is lower than six to sulfate.
There are many possible reactions. Here is one possible example:
Na2 S2O3 + 4Br2 + 5 H2O = 2NaHSO4 + 8 HBr
A spring scale is a type of weighing scale, it works because it uses a spring that is proportional to the distance that the spring is from its resting position.
